current - cnx

OpenStax-CNX module: m42341 1

Current∗

OpenStax College

This work is produced by OpenStax-CNX and licensed under the

Creative Commons Attribution License 3.0†

Abstract

• Dene electric current, ampere, and drift velocity• Describe the direction of charge ow in conventional current.• Use drift velocity to calculate current and vice versa.

1 Electric Current

Electric current is dened to be the rate at which charge ows. A large current, such as that used to starta truck engine, moves a large amount of charge in a small time, whereas a small current, such as that usedto operate a hand-held calculator, moves a small amount of charge over a long period of time. In equationform, electric currentI is dened to be

I =∆Q∆t

, (1)

where ∆Q is the amount of charge passing through a given area in time ∆t. (As in previous chapters, initialtime is often taken to be zero, in which case ∆t = t.) (See Figure 1.) The SI unit for current is the ampere(A), named for the French physicist André-Marie Ampère (17751836). Since I = ∆Q/∆t, we see that anampere is one coulomb per second:

1 A = 1 C/s (2)

Not only are fuses and circuit breakers rated in amperes (or amps), so are many electrical appliances.

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Figure 1: The rate of ow of charge is current. An ampere is the ow of one coulomb through an areain one second.

Example 1: Calculating Currents: Current in a Truck Battery and a Handheld Cal-culator(a) What is the current involved when a truck battery sets in motion 720 C of charge in 4.00 swhile starting an engine? (b) How long does it take 1.00 C of charge to ow through a handheldcalculator if a 0.300-mA current is owing?

StrategyWe can use the denition of current in the equation I = ∆Q/∆t to nd the current in part (a),

since charge and time are given. In part (b), we rearrange the denition of current and use thegiven values of charge and current to nd the time required.

Solution for (a)Entering the given values for charge and time into the denition of current gives

I = ∆Q∆t = 720 C

4.00 s = 180 C/s

= 180 A.(3)

Discussion for (a)This large value for current illustrates the fact that a large charge is moved in a small amount

of time. The currents in these starter motors are fairly large because large frictional forces needto be overcome when setting something in motion.

Solution for (b)Solving the relationship I = ∆Q/∆t for time ∆t, and entering the known values for charge and

current gives

∆t = ∆QI = 1.00 C

0.300× 10−3 C/s

= 3.33 × 103 s.(4)

Discussion for (b)This time is slightly less than an hour. The small current used by the hand-held calculator

takes a much longer time to move a smaller charge than the large current of the truck starter. Sowhy can we operate our calculators only seconds after turning them on? It's because calculators



require very little energy. Such small current and energy demands allow handheld calculators tooperate from solar cells or to get many hours of use out of small batteries. Remember, calculatorsdo not have moving parts in the same way that a truck engine has with cylinders and pistons, sothe technology requires smaller currents.

Figure 2 shows a simple circuit and the standard schematic representation of a battery, conducting path, andload (a resistor). Schematics are very useful in visualizing the main features of a circuit. A single schematiccan represent a wide variety of situations. The schematic in Figure 2 (b), for example, can represent anythingfrom a truck battery connected to a headlight lighting the street in front of the truck to a small batteryconnected to a penlight lighting a keyhole in a door. Such schematics are useful because the analysis is thesame for a wide variety of situations. We need to understand a few schematics to apply the concepts andanalysis to many more situations.

Figure 2: (a) A simple electric circuit. A closed path for current to ow through is supplied byconducting wires connecting a load to the terminals of a battery. (b) In this schematic, the battery isrepresented by the two parallel red lines, conducting wires are shown as straight lines, and the zigzagrepresents the load. The schematic represents a wide variety of similar circuits.

Note that the direction of current ow in Figure 2 is from positive to negative. The direction of conven-



tional current is the direction that positive charge would ow. Depending on the situation, positive charges,negative charges, or both may move. In metal wires, for example, current is carried by electronsthat is,negative charges move. In ionic solutions, such as salt water, both positive and negative charges move. Thisis also true in nerve cells. A Van de Graa generator used for nuclear research can produce a current ofpure positive charges, such as protons. Figure 3 illustrates the movement of charged particles that composea current. The fact that conventional current is taken to be in the direction that positive charge would owcan be traced back to American politician and scientist Benjamin Franklin in the 1700s. He named thetype of charge associated with electrons negative, long before they were known to carry current in so manysituations. Franklin, in fact, was totally unaware of the small-scale structure of electricity.

It is important to realize that there is an electric eld in conductors responsible for producing the current,as illustrated in Figure 3. Unlike static electricity, where a conductor in equilibrium cannot have an electriceld in it, conductors carrying a current have an electric eld and are not in static equilibrium. An electriceld is needed to supply energy to move the charges.

: Find a straw and little peas that can move freely in the straw. Place the straw at on a tableand ll the straw with peas. When you pop one pea in at one end, a dierent pea should pop outthe other end. This demonstration is an analogy for an electric current. Identify what compares tothe electrons and what compares to the supply of energy. What other analogies can you nd foran electric current?

Note that the ow of peas is based on the peas physically bumping into each other; electrons owdue to mutually repulsive electrostatic forces.



Figure 3: Current I is the rate at which charge moves through an area A, such as the cross-section ofa wire. Conventional current is dened to move in the direction of the electric eld. (a) Positive chargesmove in the direction of the electric eld and the same direction as conventional current. (b) Negativecharges move in the direction opposite to the electric eld. Conventional current is in the directionopposite to the movement of negative charge. The ow of electrons is sometimes referred to as electronicow.

Example 2: Calculating the Number of Electrons that Move through a CalculatorIf the 0.300-mA current through the calculator mentioned in the Example 1 (Calculating Currents:Current in a Truck Battery and a Handheld Calculator) example is carried by electrons, how manyelectrons per second pass through it?

StrategyThe current calculated in the previous example was dened for the ow of positive charge. For

electrons, the magnitude is the same, but the sign is opposite, Ielectrons = − 0.300×10 −3 C/s .Since each electron ( e− ) has a charge of −− 1.60× 10−19 C,we can convert the current in coulombs per second to electrons per second.

SolutionStarting with the denition of current, we have

Ielectrons =∆Qelectrons

∆t=−− 0.300 × 10−3 C

s. (5)



We divide this by the charge per electron, so that

e

s = −−0.300× 10−−3 Cs × 1 e

−−1.60× 10−19 C

= 1.88 × 1015 e

s .(6)

DiscussionThere are so many charged particles moving, even in small currents, that individual charges are

not noticed, just as individual water molecules are not noticed in water ow. Even more amazingis that they do not always keep moving forward like soldiers in a parade. Rather they are like acrowd of people with movement in dierent directions but a general trend to move forward. Thereare lots of collisions with atoms in the metal wire and, of course, with other electrons.

2 Drift Velocity

Electrical signals are known to move very rapidly. Telephone conversations carried by currents in wires coverlarge distances without noticeable delays. Lights come on as soon as a switch is icked. Most electricalsignals carried by currents travel at speeds on the order of 108 m/s, a signicant fraction of the speed oflight. Interestingly, the individual charges that make up the current move much more slowly on average,typically drifting at speeds on the order of 10−4 m/s. How do we reconcile these two speeds, and what doesit tell us about standard conductors?

The high speed of electrical signals results from the fact that the force between charges acts rapidly at adistance. Thus, when a free charge is forced into a wire, as in Figure 4, the incoming charge pushes othercharges ahead of it, which in turn push on charges farther down the line. The density of charge in a systemcannot easily be increased, and so the signal is passed on rapidly. The resulting electrical shock wave movesthrough the system at nearly the speed of light. To be precise, this rapidly moving signal or shock wave isa rapidly propagating change in electric eld.

Figure 4: When charged particles are forced into this volume of a conductor, an equal number arequickly forced to leave. The repulsion between like charges makes it dicult to increase the number ofcharges in a volume. Thus, as one charge enters, another leaves almost immediately, carrying the signalrapidly forward.

Good conductors have large numbers of free charges in them. In metals, the free charges are free electrons.Figure 5 shows how free electrons move through an ordinary conductor. The distance that an individualelectron can move between collisions with atoms or other electrons is quite small. The electron paths thusappear nearly random, like the motion of atoms in a gas. But there is an electric eld in the conductor that



causes the electrons to drift in the direction shown (opposite to the eld, since they are negative). The driftvelocity vd is the average velocity of the free charges. Drift velocity is quite small, since there are so manyfree charges. If we have an estimate of the density of free electrons in a conductor, we can calculate the driftvelocity for a given current. The larger the density, the lower the velocity required for a given current.

Figure 5: Free electrons moving in a conductor make many collisions with other electrons and atoms.The path of one electron is shown. The average velocity of the free charges is called the drift velocity,vd, and it is in the direction opposite to the electric eld for electrons. The collisions normally transferenergy to the conductor, requiring a constant supply of energy to maintain a steady current.

: Good electrical conductors are often good heat conductors, too. This is because large numbersof free electrons can carry electrical current and can transport thermal energy.

The free-electron collisions transfer energy to the atoms of the conductor. The electric eld does workin moving the electrons through a distance, but that work does not increase the kinetic energy (nor speed,therefore) of the electrons. The work is transferred to the conductor's atoms, possibly increasing temperature.Thus a continuous power input is required to keep a current owing. An exception, of course, is found insuperconductors, for reasons we shall explore in a later chapter. Superconductors can have a steady currentwithout a continual supply of energya great energy savings. In contrast, the supply of energy can beuseful, such as in a lightbulb lament. The supply of energy is necessary to increase the temperature of thetungsten lament, so that the lament glows.

: Find a lightbulb with a lament. Look carefully at the lament and describe its structure. Towhat points is the lament connected?

We can obtain an expression for the relationship between current and drift velocity by considering the numberof free charges in a segment of wire, as illustrated in Figure 6. The number of free charges per unit volume isgiven the symbol n and depends on the material. The shaded segment has a volume Ax, so that the number



of free charges in it is nAx. The charge ∆Q in this segment is thus qnAx, where q is the amount of chargeon each carrier. (Recall that for electrons, q is −1.60 × 10−19 C.) Current is charge moved per unit time;thus, if all the original charges move out of this segment in time ∆t, the current is

I =∆Q∆t

=qnAx

∆t. (7)

Note that x/∆t is the magnitude of the drift velocity, vd, since the charges move an average distance x ina time ∆t. Rearranging terms gives

I = nqAvd, (8)

where I is the current through a wire of cross-sectional area A made of a material with a free charge densityn. The carriers of the current each have charge q and move with a drift velocity of magnitude vd.

Figure 6: All the charges in the shaded volume of this wire move out in a time t, having a drift velocityof magnitude vd = x/t. See text for further discussion.

Note that simple drift velocity is not the entire story. The speed of an electron is much greater than itsdrift velocity. In addition, not all of the electrons in a conductor can move freely, and those that do mightmove somewhat faster or slower than the drift velocity. So what do we mean by free electrons? Atoms in ametallic conductor are packed in the form of a lattice structure. Some electrons are far enough away fromthe atomic nuclei that they do not experience the attraction of the nuclei as much as the inner electrons do.These are the free electrons. They are not bound to a single atom but can instead move freely among theatoms in a sea of electrons. These free electrons respond by accelerating when an electric eld is applied.Of course as they move they collide with the atoms in the lattice and other electrons, generating thermalenergy, and the conductor gets warmer. In an insulator, the organization of the atoms and the structure donot allow for such free electrons.

Example 3: Calculating Drift Velocity in a Common WireCalculate the drift velocity of electrons in a 12-gauge copper wire (which has a diameter of 2.053mm) carrying a 20.0-A current, given that there is one free electron per copper atom. (Householdwiring often contains 12-gauge copper wire, and the maximum current allowed in such wire isusually 20 A.) The density of copper is 8.80× 103 kg/m

3.

StrategyWe can calculate the drift velocity using the equation I = nqAvd. The current I = 20.0A is

given, and q = − −1.60 × 10−−19C is the charge of an electron. We can calculate the area of a



cross-section of the wire using the formula A = πr2, where r is one-half the given diameter, 2.053mm. We are given the density of copper, 8.80 × 103 kg/m

3, and the periodic table shows that

the atomic mass of copper is 63.54 g/mol. We can use these two quantities along with Avogadro'snumber, 6.02× 1023 atoms/mol, to determine n, the number of free electrons per cubic meter.

SolutionFirst, calculate the density of free electrons in copper. There is one free electron per copper

atom. Therefore, is the same as the number of copper atoms per m3. We can now nd n as follows:

n = 1 e−

atom× 6. 02 × 10 23 atoms

mol× 1mol

63 . 54 g× 1000 g

kg× 8.80 × 10 3 kg

1 m 3

= 8 . 342 × 10 28 e− /m 3.(9)

The cross-sectional area of the wire is

A = πr2

= π(

2.053× 10 −3 m2

)2

= 3.310 × 10 6 m 2.

(10)

Rearranging I = nqAvd to isolate drift velocity gives

vd = InqA

= 20.0 A

( 8. 342 × 10 28 /m 3) ( 1 . 60 × 10 19 C ) ( 3. 310 × 10 6 m 2)

= 4 . 53 × 10 4 m/s.

(11)

DiscussionThe minus sign indicates that the negative charges are moving in the direction opposite to

conventional current. The small value for drift velocity (on the order of 10−4 m/s) conrms thatthe signal moves on the order of 1012 times faster (about 108 m/s) than the charges that carry it.

3 Section Summary

• Electric current I is the rate at which charge ows, given by

I =∆Q∆t

, (12)

where ∆Q is the amount of charge passing through an area in time ∆t.• The direction of conventional current is taken as the direction in which positive charge moves.• The SI unit for current is the ampere (A), where 1 A = 1 C/s.• Current is the ow of free charges, such as electrons and ions.• Drift velocity vd is the average speed at which these charges move.• Current I is proportional to drift velocity vd, as expressed in the relationship I = nqAvd. Here, I is

the current through a wire of cross-sectional area A. The wire's material has a free-charge density n,and each carrier has charge q and a drift velocity vd.

• Electrical signals travel at speeds about 1012 times greater than the drift velocity of free electrons.



4 Conceptual Questions

Exercise 1Can a wire carry a current and still be neutralthat is, have a total charge of zero? Explain.

Exercise 2Car batteries are rated in ampere-hours (A · h). To what physical quantity do ampere-hourscorrespond (voltage, charge, . . .), and what relationship do ampere-hours have to energy content?

Exercise 3If two dierent wires having identical cross-sectional areas carry the same current, will the driftvelocity be higher or lower in the better conductor? Explain in terms of the equation vd = I

nqA ,by considering how the density of charge carriers n relates to whether or not a material is a goodconductor.

Exercise 4Why are two conducting paths from a voltage source to an electrical device needed to operate thedevice?

Exercise 5In cars, one battery terminal is connected to the metal body. How does this allow a single wire tosupply current to electrical devices rather than two wires?

Exercise 6Why isn't a bird sitting on a high-voltage power line electrocuted? Contrast this with the situationin which a large bird hits two wires simultaneously with its wings.

5 Problems & Exercises

Exercise 7 (Solution on p. 13.)

What is the current in milliamperes produced by the solar cells of a pocket calculator throughwhich 4.00 C of charge passes in 4.00 h?

Exercise 8A total of 600 C of charge passes through a ashlight in 0.500 h. What is the average current?


What is the current when a typical static charge of 0.250 µC moves from your nger to a metaldoorknob in 1.00 µs?Exercise 10Find the current when 2.00 nC jumps between your comb and hair over a 0.500 - µs time interval.


A large lightning bolt had a 20,000-A current and moved 30.0 C of charge. What was its duration?

Exercise 12The 200-A current through a spark plug moves 0.300 mC of charge. How long does the spark last?


(a) A debrillator sends a 6.00-A current through the chest of a patient by applying a 10,000-Vpotential as in the gure below. What is the resistance of the path? (b) The debrillator paddlesmake contact with the patient through a conducting gel that greatly reduces the path resistance.Discuss the diculties that would ensue if a larger voltage were used to produce the same currentthrough the patient, but with the path having perhaps 50 times the resistance. (Hint: The currentmust be about the same, so a higher voltage would imply greater power. Use this equation forpower: P = I2R.)



Figure 7: The capacitor in a debrillation unit drives a current through the heart of a patient.

Exercise 14During open-heart surgery, a debrillator can be used to bring a patient out of cardiac arrest. Theresistance of the path is 500 Ω and a 10.0-mA current is needed. What voltage should be applied?


(a) A debrillator passes 12.0 A of current through the torso of a person for 0.0100 s. How muchcharge moves? (b) How many electrons pass through the wires connected to the patient? (Seegure two problems earlier.)

Exercise 16A clock battery wears out after moving 10,000 C of charge through the clock at a rate of 0.500mA. (a) How long did the clock run? (b) How many electrons per second owed?


The batteries of a submerged non-nuclear submarine supply 1000 A at full speed ahead. How longdoes it take to move Avogadro's number (6.02× 1023) of electrons at this rate?

Exercise 18Electron guns are used in X-ray tubes. The electrons are accelerated through a relatively largevoltage and directed onto a metal target, producing X-rays. (a) How many electrons per secondstrike the target if the current is 0.500 mA? (b) What charge strikes the target in 0.750 s?


A large cyclotron directs a beam of He++ nuclei onto a target with a beam current of 0.250 mA.(a) How many He++ nuclei per second is this? (b) How long does it take for 1.00 C to strike thetarget? (c) How long before 1.00 mol of He++ nuclei strike the target?



Exercise 20Repeat the above example on Example 3 (Calculating Drift Velocity in a Common Wire), but fora wire made of silver and given there is one free electron per silver atom.


Using the results of the above example on Example 3 (Calculating Drift Velocity in a CommonWire), nd the drift velocity in a copper wire of twice the diameter and carrying 20.0 A.

Exercise 22A 14-gauge copper wire has a diameter of 1.628 mm. What magnitude current ows when the driftvelocity is 1.00 mm/s? (See above example on Example 3 (Calculating Drift Velocity in a CommonWire) for useful information.)


SPEAR, a storage ring about 72.0 m in diameter at the Stanford Linear Accelerator (closed in2009), has a 20.0-A circulating beam of electrons that are moving at nearly the speed of light. (SeeFigure 8.) How many electrons are in the beam?

Figure 8: Electrons circulating in the storage ring called SPEAR constitute a 20.0-A current. Becausethey travel close to the speed of light, each electron completes many orbits in each second.



Solutions to Exercises in this Module

Solution to Exercise (p. 10)0.278 mASolution to Exercise (p. 10)0.250 ASolution to Exercise (p. 10)1.50msSolution to Exercise (p. 10)(a) 1.67 kΩ

(b) If a 50 times larger resistance existed, keeping the current about the same, the power would beincreased by a factor of about 50 (based on the equation P = I2R), causing much more energy to betransferred to the skin, which could cause serious burns. The gel used reduces the resistance, and thereforereduces the power transferred to the skin.Solution to Exercise (p. 11)(a) 0.120 C

(b) 7.50× 1017 electronsSolution to Exercise (p. 11)96.3 sSolution to Exercise (p. 11)(a) 7.81× 1014 He++ nuclei/s

(b) 4.00× 103 s(c) 7.71× 108 s

Solution to Exercise (p. 12)−1.13× 10−4 m/sSolution to Exercise (p. 12)9.42× 1013 electrons

Glossary

Denition 1: electric currentthe rate at which charge ows, I = ∆Q/∆t

Denition 2: ampere(amp) the SI unit for current; 1 A = 1 C/s

Denition 3: drift velocitythe average velocity at which free charges ow in response to an electric eld


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